Class AB Power Amplifier Design

[MD25]

Hello All,

I'm trying to build a Class AB amplifier to push an audio signal through an 8ohm speaker. I've opted to go transformerless and intend to use diode biasing. I'm also using BD135 and BD136 transistors. My circuit essentially looks like the circuit below except I have VCC as +15 VDC and 0V as -15VDC. I understand that the resistors I use between the voltage rails and the transistors will drop 14.3V regardless of value as the diodes drop 0.7V (Kirchhoff's Voltage Law). The resistors will however change the amount of current going into the bases and hence will effect the amount of current flow available on the output. I'm using Multisim to simulate my design and I am only getting a power dissipation of about 60mW @ 100mA in the speaker load with the resistors set to just 10ohms. I have a function generator set to 1V @ 1kHz but I don't think that matters here. Is there something I'm missing? Am I going about this design the wrong way? If you need more info, just ask and I will provide gladly.

Kind Regards,

MD25

 

[Audioguru]

Your Q-point is not at half the supply voltage, instead it is too close to 0V.
EDIT: Your schematic and load line do not show -15V, instead they show ground.
Your circuit uses the driving signal to turn "off" the transistors but most class-AB audio amplifiers use the driving signal to turn "on" at least one of the transistors.

Here is a simple class-AB amplifier:

 

[MD25]

Your Q-point is not at half the supply voltage, instead it is too close to 0V.
EDIT: Your schematic and load line do not show -15V, instead they show ground.
Your circuit uses the driving signal to turn "off" the transistors but most class-AB audio amplifiers use the driving signal to turn "on" at least one of the transistors.

Here is a simple class-AB amplifier:

Hello there and thanks for your response. That circuit diagram is an example that I used from another website and where it shows 0V on that diagram, -15V is connected there instead on my actual circuit. The speaker however is grounded, should it be referenced to -15V?

Thanks in advance.

 

[Audioguru]

With a single-ended amplifier output then the speaker is always grounded. A bridged amplifier has two amplifiers, one of the amplifiers drives one speaker wire and the other amplifier drives the other speaker wire with opposite phase for double the voltage which produces almost 4 times the power. A bridged amplifier is used in car radios and has no negative supply voltage then each speaker wire has a DC voltage at half the power supply voltage and no DC current in the speaker so output coupling capacitors are not needed.

The circuit you posted and my circuit do not have a negative power supply voltage so they have an output at half the supply voltage. An output coupling capacitor is needed as I show on mine so that there is no DC current in the speaker. With a plus and minus power supply then the output is already at 0VDC so an output coupling capacitor is not needed.

Do you understand that the output DC voltage must be at half the supply voltage (0V when the supply is plus and minus) so that the output can swing equally up and down?

Do you understand that bootstrapping increases the positive output voltage swing, increases voltage gain and reduces distortion?

 

[Harald Kapp]

With a single-ended amplifier output then the speaker is always grounded.

And so is the input.

The circuit you posted and my circuit do not have a negative power supply voltage

The 0V rail is at -15 V according to the op's post #1.
@MD25 : It is misleading to show a circuit diagram that does not represent the actual circuit. You could have avoided this misunderstanding by uploading a hardcopy of your actual multisim schematic with correct voltages instead. I take it your circuit looks like this (sorry for the wrong transistor types, I just didn't have the matching models at hand):


This circuit has a voltage gain of ~1. The output voltage therefore is also 1 V (peak) for an input voltage of 1 V (peak) which is equivalent to 0.707 V (rms).
The power dissipation in an 8 Ω load is then P = V²/R = 0.5 V²/8 Ω = 62.5 mW which is in good agreement with your simulation.
This kind of stage with a voltage gain of ~1 is the power stage to drive a high current into a low impedance load. You need pre-amplifiers to achieve the required voltage gain such that Vin is a signal with a voltage swing between e.g. +14 V ... - 14 V (from a +- 15 V supply).

 

[Audioguru]

I think MD25 does not know that the output transistors are emitter-followers with no amplification of the voltage swing.
Why did he use resistors that are only 10 ohms? Then the current in the resistors is about 14.3V/10 ohms= 1.43A!
The amplifier I posted has an input of 0.48V peak and an output of 11V peak then its voltage gain is 11/0.48= 23 times. The current in its resistors that feed the diodes is only 11.3V/440 ohms= 26mA.

 

[AnalogKid]

I understand that the resistors I use between the voltage rails and the transistors will drop 14.3V regardless of value as the diodes drop 0.7V (Kirchhoff's Voltage Law).

Nope. What you say is true only if there is no input signal or the input signal is passing through Vcc/2 V. As the input signal changes, it drives the bias string moves up and down, and thus the output.

ak